c-变量的初始化列表

是否可以创建变量的initializer_list,例如函数参数(例如,功能测试)？

下面的代码有效,并且Clang和GCC都没有抱怨任何事情,但是我只想确保这是可以的.

#include <iostream>
#include <initializer_list>

template <class T>
struct array
{
    T *ptr;
    size_t len;

    array() { clear(); }
    array( T *p, size_t l ) { assign(p,l); }

    inline void clear() { ptr=nullptr; len=0; }
    inline void assign( T *p, size_t l ) { ptr=p; len=l; }

    inline T& operator[] ( size_t i ) const { return ptr[i]; }
};

template <class T>
inline array<const T> wrap( const std::initializer_list<T>& lst )
    { return array<const T>( lst.begin(), lst.size() ); }

void test( int a, int b, int c )
{
    auto ar = wrap({a,b,c});
    std::cout<< ar[2] << std::endl;
}

int main()
{
    auto a = wrap({1,2,3});
    std::cout<< a[2] << std::endl;

    test(1,2,3);
}

附带问题；如果我尝试在测试中返回包装好的数组,则初始化列表{a,b,c}会超出范围,而我返回的数组将无效-这对吗？

解决方法:

 auto ar = wrap({a,b,c}); 

这将创建一个类型为int [3]的临时数组,然后绑定一个initializer_list< int>.对该数组进行调用,然后调用wrap创建一个数组const int.指的是数组.

在表达式的末尾,数组被销毁,而数组const int被保留.带有悬空的指针,因此这是未定义的行为：

 std::cout<< ar[2] << std::endl;

这也适用于main中的代码,变量a包含一个悬空指针,而a [2]是未定义的行为.

您可以通过将int数组替换为分配内存的类型数组来验证这一点,以便valgrind或asan将注意到该错误：

using V = std::vector<int>;
auto a = wrap({V{1}, V{2}, V{3}});
std::cout<< a[2].front() << std::endl;

现在a [2]是std :: vector< int>对象,但是尝试访问其front()成员会导致程序中止：

==28356==ERROR: AddressSanitizer: heap-use-after-free on address 0x60200000efb0 at pc 0x000000401205 bp 0x7fffa46f2900 sp 0x7fffa46f28f8
READ of size 4 at 0x60200000efb0 thread T0
    #0 0x401204 in main /tmp/il.cc:28
    #1 0x3236e21d64 in __libc_start_main (/lib64/libc.so.6+0x3236e21d64)
    #2 0x400ec8  (/tmp/a.out+0x400ec8)
...

或使用valgrind：

==28364== Invalid read of size 4
==28364==    at 0x400C72: main (il.cc:28)
==28364==  Address 0x51dfd20 is 0 bytes inside a block of size 4 free'd
==28364==    at 0x4A07991: operator delete(void*) (vg_replace_malloc.c:502)
==28364==    by 0x4013BF: __gnu_cxx::new_allocator<int>::deallocate(int*, unsigned long) (new_allocator.h:110)
==28364==    by 0x4012F8: std::allocator_traits<std::allocator<int> >::deallocate(std::allocator<int>&, int*, unsigned long) (alloc_traits.h:386)
==28364==    by 0x4011B1: std::_Vector_base<int, std::allocator<int> >::_M_deallocate(int*, unsigned long) (stl_vector.h:178)
==28364==    by 0x40102A: std::_Vector_base<int, std::allocator<int> >::~_Vector_base() (stl_vector.h:160)
==28364==    by 0x400EC4: std::vector<int, std::allocator<int> >::~vector() (stl_vector.h:425)
==28364==    by 0x400C2A: main (il.cc:27)

Side question; if I tried to return my wrapped array in test, the initializer list {a,b,c} would get out of scope, and the array I’m returning would be invalid — is that correct?

它已经超出范围,并且即使在返回之前也已经无效.

标签：c,c11,initializer-list
来源： https://codeday.me/bug/20191013/1907197.html